E- for the reaction Fe - Zn2- Zn - Fe2- is -0-35 V- The given reaction is-

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Byju's Answer

Standard VII

Chemistry

EMF

E∘ for the re...

Question

$E^{\circ}$ for the reaction $F e + Z n^{2 +} \to Z n + F e^{2 +}$ is -0.35 V. The given reaction is:

Feasible

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Not feasible

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In equilibrium

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None

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Solution

The correct option is B Not feasible
Since EMF of the cell is negative i.e., free energy change will be positive so cell reaction will not be feasible.

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Similar questions

Q. The standard reduction potentials

E^{⊝}

for the half reactions are follows :

Z n \to Z n^{2 +} + 2 e^{-}; E^{⊝} = + 0.76 V

F e \to F e^{2 +} + 2 e^{-}; E^{⊝} = 0.41 V

The EMF for the cell reaction

F e^{2 +} Z n \to Z n^{2 +} + F e

is:

Q. The standard oxidation potentials,

E^{\circ}

, for the half reactions are as,

Z n \to Z n^{2 +} + 2 e^{-}; E^{\circ} = + 0.76 v o l t

F e \to F e^{2 +} + 2 e^{-}; E^{\circ} = + 0.41 v o l t

The emf of the cell,

F e^{2 +} + Z n \to Z n^{2 +} + F e

is:

The standard reduction potentials, $E^{\circ}$ , for the half-reactions are as $Z n = Z n^{2 +} + 2 e$ , $E^{\circ}$ = + 0.76 V and $F e = F e^{2 +} + 2 e$ , $E^{\circ}$ = + 0.41 V. The e.m.f. for the cell reaction, $F e^{2 +} + Z n = Z n^{2 +} + F e$ is